nFeCl3=0,4.0,2=0,08(mol)
PTHH: FeCl3+3NaOH-> Fe(OH)3+3NaCl
(mol) 0,08 ->0,24
CM ddNaOH=\(\dfrac{0,24}{0,2}=1,2M\)
\(n_{F\text{e}Cl_3}=0.4x0.2=0.08\)
Pt: \(F\text{e}Cl_3+3NaOH\rightarrow F\text{e}\left(OH\right)_3+3NaCl\)
0.08 \(-\)2.4
Từ pt ta có: \(n_{NaOH}=2.4mol\)
\(C_M=\dfrac{2.4}{0.2}=12M\)